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I saw something like this:

To extend the proof to all random variables, let $\widehat{X}$ be the random variable $X$ conditioned on $X$ being non-negative. The distribution of $\widehat{X}$ is given by $$ F_{\widehat{X}}(x) = \frac {F_X(x)- (1 - F_X(0))} {\mathbb P[X \ge 0]}, \quad x \ge 0, $$ so that
$$dF_{X}(x) =\frac {dF(x)} {\mathbb P[X > 0]}.$$

The definition of $F_{\widehat{X}}(x)$ doesn't quite make sense for me. For example, when $x = 0$, we might have $F_{\widehat{X}}(x) < 0$, which does not match the definition of a distribution function. So, how should I interpret this? The about text is excerpted from this paper.

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I think, you're right: $$ F_{\hat X}(0) = \frac{2F_X(x) - 1}{\Bbb P[X\geq 0]} $$ which can easily be negative. I would rather condition as $$ \Bbb P[\hat X \in A] := \Bbb P[X\in A|X\geq 0], $$ but I'm not sure that's what they mean cause in this case you get $$ F_{\hat X}(x) = \frac{F_X(x) - F_X(0)}{\Bbb P[X\geq 0]}, \quad x\geq 0 $$ which is indeed a CDF provided the denominator is non-zero. So perhaps it's a kinda typo in their case as they use $1-F_X(0)$ in place of $F_X(0)$.

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Seems that this is the only possible explanation. –  ablmf Mar 25 '13 at 18:15
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